DateDiff("d", #5/1/2009#, #6/4/2009#) will return the number of days
(I'm assuming you want days).
Assuming your data is actually in table fields;
DateDiff("d", [Field1], [Field2])
If you want to include the second field in the returned result you'll
have to add one;
DateDiff("d", [Field1], [Field2]) + 1
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